**Exercise 6.3**

Permutations and combinations

**Question and Answers**

**Class 11 – Maths**

Permutations and combinations

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Permutations and Combinations |

Chapter No. | Chapter 6 |

Exercise | Exercise 6.3 |

Category | Class 11 Maths NCERT Solutions |

**Question 1 How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?**

**Answer **Total number of digits possible for choosing = 9

Number of places for which a digit has to be taken = 3

As there is no repetition allowed , Number of permutations

= 9×8×7=504

**Question 2 How many 4-digit numbers are there with no digit repeated?**

**Answer **The thousands place of the 4-digit number can be filled with any digit from 1 to 9 and 0 is not included. Thus, number of ways in which thousands place is filled is 9.

The hundreds, tens and units place can be filled with any digit from 0 to 9.

Since, the digits cannot be repeated and thousands place is already occupied by the digit. The hundreds, tens and units place can be filled by remaining 9 digits.

Thus, there are permutations of 9 different digits taken 3 at a time.

Number of 3-digit numbers =

=9×8×7=504

Number of 4-digit numbers is 9 × 504 = 4536

**Question 3 How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?**

**Answer** The units place can be filled by any of the digits 2, 4 or 6. Hence, there are 3 ways.

Digits cannot be repeated and units place is already occupied so the hundreds and tens place can be occupied by remaining 5 digits.

Thus, number of ways of filling hundreds and tens place =

= 20

Therefore, the total number of permutations =3 × 20=60

**Question 4 Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?**

**Answer **Total number of digits possible for choosing = 5

Number of places for which a digit has to be taken = 4

There are permutations of 5 different things taken 4 at a time.

Thus, number of 4 digit numbers =

= 1×2×3×4×5 = 120

Out of 1, 2, 3, 4, 5, we know that even numbers end either by 2 or 4.Thus, ways in which units place can be filled is 2.

Since, repetition is not allowed, units place is already occupied by a digit and remaining vacant places can be filled by remaining 4 digits.

Thus, number of ways in which remaining places can be filled =

=4×3×2×1 = 24

Number of even numbers = 24×2=48

**Question 5 From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?**

**Answer** In a committee of 8 persons, a chairman and vice chairman are selected in such away that one person can hold only one position.Thus, ways of choosing a chairman and vice chairman is permutation of 8 objects taken 2 at a time.

Therefore, number of ways

**Question 6 Find n if ^{n-1}P_{3 }: ^{n}P_{3} = 1: 9.**

**Answer**

**Question 7 Find r if **

**(i) ^{5}P_{r} = 2^{6}P_{r-1} **

(7−r) (6−r) =12

⇒42− 6r− 7r + r = 12

⇒r^{2}−13r + 30 = 0

⇒r^{2}− 3r + 10r + 30

⇒r(r−3)−10(r−3)=0

⇒(r−3)(r−10)=0

⇒ ( r – 3) = 0 or (r−10) = 0

⇒r = 3 or r = 10

∴0 ⩽ r ⩽ 5

∴ r=3

**(ii) ^{5}P_{r} = ^{6}P_{r-1}**

(7−r) (6−r) = 6

= 42−7r−6r+r^{2}−6=0

= r^{2}−13r + 36=0

= r^{2}− 4r − 9r (r−4)=0

= (r−4) (r−9)=0

= (r−4 )(r−9)=0

= (r−4)=0 or (r−9)=0

= r=4 or r=9

Hence, r ≠ 9

∴r = 4

**Question 8 How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?**

**Answer **The number of different letters in the given word is 8.

Thus, the number of words than can be formed without repetition is number of permutations of 8 different objects taken 8 at a time = ^{8}P_{8 }= 8!

Therefore, number of words formed = 8! = 40320.

**Question 9 How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. **

**(i) 4 letters are used at a time,**

**(i)** Number of different letters in the given word = 6

Number of 4-letter words that can be formed from the letters of the given word without repetition is permutations of 6 different objects taken 4 at a time.

Therefore, Number of 4 letter words that can be formed

** (ii) all letters are used at a time,**

**(ii)** Number of different letters in the given word = 6

Words that can be formed using all the letters of the given word is permutation of 6 different objects taken 6 at a time.

^{6}P_{6} = 6!

Therefore, number of words that can be formed is

= 6! = 6×5×4×3×2×1 = 720

**(iii) all letters are used but first letter is a vowel?**

**(iii)** There are two different vowels in the word MONDAY which occupies the rightmost place of the words formed. Hence, there are 2 ways.

Since, it is without repetition and the rightmost place is occupied, the remaining five vacant places can be filled by 5 different letters. Hence, 5! Ways.

Therefore, number of words that can be formed = 5!×2 = 120×2 = 240

**Question 10 In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?**

**Answer** Total number of letters in MISSISSIPPI =11

I appears 4 times, S appears 4 times, P appears 2 times and M appears one time in the given word

Thus, number of permutations of the given word

We take that 4 I’s come together, and they are treated as 1 letter, ∴ Total number of letters=11 – 4 + 1 = 8

Number of permutations = 8! / 4! 2! = 840 ways

Therefore, total number of permutations where four I’s don’t come together = 34650-840=33810

**Question 11 In how many ways can the letters of the word PERMUTATIONS be arranged if the**

** (i) words start with P and end with S,**

**(i)** Total number of letters in PERMUTATIONS =12

The only repeated letter is T; 2times whereas the first and last letters of the word are fixed as P and S.Hence, 10 letters are left.

Therefore, required number of permutations =

** (ii) vowels are all together,**

(ii) There are 5 vowels in the given word, each appearing only once.

Since they have to always occur together, they are treated as a single object for the time being. This single object together with the remaining 7 objects will account for 8 objects, in total.

These 8 objects in which there are 2 T’s can be arranged in 8! / 2! ways.

Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5 !.

Therefore the required number of arrangements are :

( 8 ! × 5 ! ) / 2 ! =( 8 × 7 × 6 × 5 × 4 ×3× 2 × 120 ) / 2 = 24192008

**(iii) there are always 4 letters between P and S?**

(iii) The letters have to be arranged in such a way that there are always 4 letters between P and S. Therefore, in a way, the places of P and S are fixed. The remaining 10 letters in which there are 2 T’s can be arranged in 10! / 2! ways.

Possible places of P & S are –

P and S can take the position 1 and 6

P and S can take the position 2 and 7

P and S can take the position 3 and 8

P and S can take the position 4 and 9

P and S can take the position 5 and 10

P and S can take the position 6 and 11

P and S can take the position 7 and 12

Also P & S can be interchanged as it won’t effect the number of letters between them.

So, the letters P and S can be placed such that there are 4 letters between them in 2 × 7 =14 ways.

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